Question

Evaluate: $\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}$

Answer 1

$\frac{sin{18}^{\circ }}{cos{72}^{\circ }}=\frac{sin({90}^{\circ }-{72}^{\circ })}{cos{72}^{\circ }}=\frac{cos{72}^{\circ }}{cos{72}^{\circ }}=1$